Question

A circular garden of radius 10 m is divided into two parts by a straight line fence. Smaller part is the walking area and flowers are planted in the larger part. The fence is at a distance of 6 m from the centre of the garden. What is the walking area in $m^2$? It is given that cos 53°= $\frac{3}{5}.$

• A. $36m^2$
• B. $44.5m^2$
• C. $12m^2$
• D. $72m^2$

Answer 1

The correct option is B

$44.5m^2$

In $△ORQ,$
${10}^2$ = ${OR}^2$ + ${RQ}^2$
${10}^2$ = $6^2$ +${RQ}^2$
$\Rightarrow$ ${RQ}^2$ = 64
$\Rightarrow$ RQ = 8 m
$\therefore$ PQ = 2RQ = 16 m [OR is perpendicular bisector of PQ]

Area of $\Delta$OPQ = $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × PQ × OR
= $\frac{1}{2}$ × 16 × 6
= 48 $m^2$

Area of sector OPSQ = $\frac{\angle POQ}{{360}^{\circ }}$ × $\pi \times r^2$
In ORQ,
cos($\angle$ROQ) = $\frac{\text{adjacent side}}{\text{hypotenuse}}$
= $\frac{OR}{OQ}$
= $\frac{6}{10}$
= $\frac{3}{5}$
$\Rightarrow$ cos( $\angle$ROQ) = $\frac{3}{5}$
It is given that cos(${53}^{\circ }$) = $\frac{3}{5}$.
Hence $\angle$ROQ = ${53}^{\circ }$

$\angle$POQ = 2($\angle$ROQ) = ${106}^{\circ }$

Area of sector POQS = $\frac{POQ}{360}$ ×$\pi \times r^2$
= $\frac{{106}^{\circ }}{{360}^{\circ }}$ ×$\pi \times {10}^2$ = 92.5 $m^2$

Area of segment PRQS = Area of sector POQS – Area of $△$OPQ = 92.5 – 48 = 44.5 $m^2$